Metric Form of ##g_{μν}## - Solving a Challenge

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SUMMARY

The forum discussion centers on the metric form of the equation ##ds^2=g_{μν}(x)dx^μdx^ν= -(r^6/l^6)[1-(Ml^2/r^2)]dt^2+{1/[(r^2/l^2)-M]}dr^2+r^2dΦ^2## and the challenges associated with interpreting and manipulating it. Participants emphasize the need for context regarding the metric's origin, specifically referencing the paper found at arxiv:0909.1347. Key questions include the meaning of terms like ##∂∨μ## and ##√-g##, as well as how to derive the form of ##g^{μν}## from the given metric. The discussion highlights the importance of clarity in mathematical notation and understanding the underlying concepts of general relativity.

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  • Study the derivation of metric tensors from line elements in general relativity
  • Learn about the significance of the determinant of the metric tensor, ##g##
  • Explore the implications of the equation of motion ##∂µ[√ (−g)gµν∂ν(ϕ)]− m^2√ (−g)ϕ = 0##
  • Review the paper at arxiv:0909.1347 for deeper insights into the metric discussed
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This discussion is beneficial for physicists, mathematicians, and students specializing in theoretical physics, particularly those focusing on general relativity and tensor calculus.

hykhor0601
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TL;DR
Does anyone know how to solve this?
##ds^2=g_{μν}(x)dx^μdx^ν= -(r^6/l^6)[1-(Ml^2/r^2)]dt^2+{1/[(r^2/l^2)-M]}dr^2+r^2dΦ^2##

Does anyone know how to solve this?
 
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What do you mean by "solve"? There is nothing to "solve", there is just a line-element.
 
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Solve in what sense? It appears to be just a statement of a metric, missing a definition of ##l## and assuming ##d\Phi## is a solid angle element.
 
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Can you provide some context here? Where di this equation come from? What book course etc...?

and what have you tried so far?

We had to modify your equation using Latex to make it readable:
ds^2=gμν(x)dx^μdx^ν= -(r^6/l^6)[1-(Ml^2/r^2)]dt^2+{1/[(r^2/l^2)-M]}dr^2+r^2dΦ^2

Basically the changes were to bracket the equation on both sides with double # and to replace the gμν with g_{μν} to make them into subscripts.
 
jedishrfu said:
Can you provide some context here? Where di this equation come from? What book course etc...?

and what have you tried so far?

We had to modify your equation using Latex to make it readable:Basically the changes were to bracket the equation on both sides with double # and to replace the gμν with g_{μν} to make them into subscripts.
https://arxiv.org/pdf/0909.1347.pdf The metric is given in (9)
 
Ibix said:
Solve in what sense? It appears to be just a statement of a metric, missing a definition of ##l## and assuming ##d\Phi## is a solid angle element.
The conditions is in the article below, please help me
https://arxiv.org/pdf/0909.1347.pdf
 
hykhor0601 said:
The conditions is in the article below, please help me
https://arxiv.org/pdf/0909.1347.pdf
This still does not answer the question you have been asked about what you mean by ”solve”.
 
∂∨μ[(√-g)g∨μν∂∨ν(Φ)]-m^2√(-g)Φ=0
What is the meaning of ∂∨μ,√-g,∂∨ν(Φ) in this equation?

How to give the form of g^μν from the metric (9) in the article?(or how do i get the form of ∂∨μ[(√-g)g∨μν∂∨ν(Φ)]-m^2√(-g)Φ=0 )
https://arxiv.org/pdf/0909.1347.pdf
Please help me, thank you
 
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hykhor0601 said:
The conditions is in the article below, please help me
https://arxiv.org/pdf/0909.1347.pdf
Again, what do you mean by "solve"? The metric is not a thing you solve. What are you trying to find out?
 
  • #10
Why are you interested in this particular paper if you are not familiar with basics of general relativity?
 
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  • #11
Ibix said:
Again, what do you mean by "solve"? The metric is not a thing you solve. What are you trying to find out?
just the form of gμν that can be sub into the equation of motion
 
  • #12
hykhor0601 said:
just the form of gμν that can be sub into the equation of motion
You already have ##g_{\nu\nu}##. What is your actual problem?
 
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  • #13
Orodruin said:
You already have ##g_{\nu\nu}##. What is your actual problem?
can you write it in the form after substitude into
∂µ[√ (−g)gµν∂ν(ϕ) ]− m^2√ (−g)ϕ = 0
 
  • #14
I don't understand. You can read off the components of ##g_{\mu\nu}## from the line element and ##g## is just its determinant. Where's the difficulty?

It would also help immensely if you spent a few minutes reading the LaTeX Guide linked below the reply box. Then you could type maths in a way that we could actually read it.
 
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